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3 years ago

Is muddy water a substance?

Physics

2 answers:

sergejj [24]3 years ago

7 0

Answer: No, I wouldn't imagine so.

Explanation:

Kazeer [188]3 years ago

7 0

Muddy water is not a substance

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While in empty space, an astronaut throws a ball at a velocity of 9 m/s. What will the velocity of the ball be after it has trav

Snezhnost [94]

Answer:

1.5

Explanation:

5 0

2 years ago

You are given three pieces of wire that have different shapes (dimensions). You connect each piece of wire separately to a batte

VikaD [51]

Answer:

$R_3 < R_1 < R_2$

Explanation:

The resistance of a wire is given by:

$R=\frac{\rho L}{A}$

where

$\rho$ is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

1) The first wire has length L and cross-sectional area A. So, its resistance is:

$R_1=\frac{\rho L}{A}$

2) The second wire has length twice the first one: 2L, and same thickness, A. So its resistance is

$R_2=\frac{2\rho L}{A}$

3) The third wire has length L (as the first one), but twice cross sectional area, 2A. So, its resistance is

$R_3=\frac{\rho L}{2A}$

By comparing the three expressions, we find

$R_3 < R_1 < R_2$

So, this is the ranking of the wire from most current (least resistance) to least current (most resistance).

5 0

2 years ago

Help with these three

Elenna [48]

The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
$g = 9.81 \frac{m}{s {}^{2} }$
let's do it for this case:
$f_{g} = m \times g \\ f _{g} = 65kg \times 9.81 \frac{m}{s {}^{2} } = 637.65$
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
$f = m \times a \\ f = 65 \times 5 \frac{m}{s {}^{2} } \\ f = 325n$
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere

8 0

3 years ago

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

6 0

3 years ago

A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal

Sati [7]

(a) The ball has a final velocity vector

$\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j$

with horizontal and vertical components, respectively,

$v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}$

$v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}$

The horizontal component of the ball's velocity is constant throughout its trajectory, so $v_{x,i}=v_{x,f}$ , and the horizontal distance x that it covers after time t is

$x=v_{x,i}t=v_{x,f}t$

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

$103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s$

The vertical component of the ball's velocity at time t is

$v_{y,f}=v_{y,i}-gt$

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

$-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}$

So, the initial velocity vector is

$\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which carries an initial speed of

$\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}$

and direction θ such that

$\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}$

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height y at time t is

$y=v_{y,i}t-\dfrac12gt^2$

so that when it lands in the seats at t ≈ 6.38 s, it has a height of

$y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}$

6 0

3 years ago